The ratio between a two-digit number and the sum of the digits of that number is 4:1. If the digit in the unit place is 3 more than the digit in the ten's place, what is that number?
Answer & explanation
Correct answer: option 3
Let the ten's digit be $x$.
Then the unit's digit is $x + 3$.
So, the number is:
$10x + (x + 3) = 11x + 3$
The sum of the digits is:
$x + (x + 3) = 2x + 3$
Given:
$\frac{\text{Number}}{\text{Sum of digits}} = \frac{4}{1}$
$\frac{11x + 3}{2x + 3} = 4$
$11x + 3 = 8x + 12$
$3x = 9$
$x = 3$
Therefore:
- Ten's digit = 3
- Unit's digit = 6
Hence, the number is: 36